bug fix in function one_to_one

[techkang]

I think I found a bug in function one_to_one. Suppose there are one predict and two ground truth, the sigma table is [1,1]^T and tau table is [1,0]^T. This is an many to one case but origin code treat it as one to one case.

[ckchng]

Hi there,

Thanks for reporting this. At a quick glance, I don't quite get why is the following line fails to make sure it's a one-to-one case. Do you have some examples on your hand to further illustrate this? Thanks.

(num_qualified_sigma_candidates == 1) and (num_qualified_tau_candidates == 1)

[techkang]

According to det eval paper:

one-to-one matches: one ground truth rectangle Gi matches with a result rectangle Dj if row i of both matrices contains only one element satisfying (10) and column j of both matrices contains only one element satisfying (10). This situation is shown in figure 2a.

I think your code only checked row i of both matrices rather that checked row i and column j of both matrices.

[ckchng]

Hi there,

The first and the third lines are supposed to check all rows and columns of both matrices.

qualified_sigma_candidates = np.where(local_sigma_table[gt_id, :] > tr) num_qualified_sigma_candidates = qualified_sigma_candidates[0].shape[0] qualified_tau_candidates = np.where(local_tau_table[gt_id, :] > tp) num_qualified_tau_candidates = qualified_tau_candidates[0].shape[0]

[techkang]

Hi there,

I think your code go over every row and check elements of each row. But the correct code should be go over each row and check elements of each row. If there is one candidate, check its corresponding column.

[ckchng]

Yea I came to this conclusion as well this morning. I will do a testing and merge / rewrite this part soon. Thank you so much for your effort!